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 College Mathematics studyhard Unregistered   #1 07-23-2007, 06:05 PM Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them: If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ?????? I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas? DixieGirl Junior Member  Posts: 42 Threads: 8 Likes Received: 0 in 0 posts Likes Given: 0 Joined: Jul 2007 #2 07-24-2007, 12:44 PM studyhard Wrote:Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them: If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ?????? I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas? Did you mean 8^x = 15 8^y = 25 8^(2x+y) = ?? I couldn't figure it out either, but I posted this on another forum and they said, "since 8x = 15 ... 82x = (8x)2 = 152. 82x+y = 82x*8y = 152*25" Hope this helps!:o wannabeit Senior Member    Posts: 277 Threads: 8 Likes Received: 0 in 0 posts Likes Given: 0 Joined: Aug 2006 #3 07-24-2007, 01:49 PM studyhard Wrote:Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them: If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ?????? I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas? I think (x =15 and (y)=25, therefore 8 x (2 x 15+ 25) (30 +25) = 55 8 x 55 = 440 BSIT- Excelsior 2010, MCSE (2003), CCNP, Project+ sgloer Administrator       Posts: 975 Threads: 139 Likes Received: 137 in 48 posts Likes Given: 3 Joined: Jul 2017 #4 07-24-2007, 02:20 PM (This post was last modified: 07-24-2007, 02:22 PM by sgloer.) I believe this is how it works. There's a rule involving multiplying exponents with the same base number, so: 8^(2x + y) is the same as 8^(2x) + 8^(y) which is the same as 8^x + 8^x + 8^y = 15 + 15 + 25 = 55 If you would provide the correct answer, that would help with reverse engineering the rule/method, because I might be remembering the rule incorrectly. -- Steve Webmaster, InstantCert.com tryingtesc Junior Member  Posts: 5 Threads: 1 Likes Received: 0 in 0 posts Likes Given: 0 Joined: Jul 2007 #5 07-27-2007, 02:45 PM Ok...There are two ways to solve this problem...One is the use of Logs to find an unknown exponent. For example: 8^x=15 = log 15 base 8 = x (re-write in log form) =log 15 / log 8 (divide) =1.3022968652 (this is the "x" exponent) 8^y=25 =log 25 base 8 = y =log 25 / log 8 = 1.54795206326 (this is the "y" exponent) Plug your now Known exponents into your equation: 8^2x+y = 8^2(1.3022968652)+(1.54795206326) = 5624.9999 = 5625 Second Way: They made this easy because of the same base 8. 8^x=15 , 8^y=25, 8^2x+y=?? 8^2x+y = 8^2x * 8^y (re-write by seperating exponents) 8^y is the easy part, we know that is 25 8^2x is not the same of course as 8^x (which we know is 15) so what makes it different? Yes the "2". Well you squared the left side of the equation so you have to square the right side: 8^2x = 15^2. Now you know 8^2x = 225 (15 squared). So put the two together: 8^2x * 8^y is now the same as 225*25 = 5625 ^ = raised to that exponent * = multiplication I hope this helps. It is a tough problem to explain on a message board. Best regards, Jason tryingtesc Junior Member  Posts: 5 Threads: 1 Likes Received: 0 in 0 posts Likes Given: 0 Joined: Jul 2007 #6 07-27-2007, 06:11 PM (This post was last modified: 07-27-2007, 06:15 PM by tryingtesc.) sgloer Wrote:I believe this is how it works. There's a rule involving multiplying exponents with the same base number, so: 8^(2x + y) is the same as 8^(2x) + 8^(y) which is the same as 8^x + 8^x + 8^y = 15 + 15 + 25 = 55 If you would provide the correct answer, that would help with reverse engineering the rule/method, because I might be remembering the rule incorrectly. You have the problem set up nicely, however, you should multiply in rows 2, 3, and 4 instead of add. You can "split" the exponent, but you then must multiply. For example: If you had x^2+3; that would equal (x^2)(x^3), the bases are the same and in multiplication you would simply add the exponents making it x^5. You could not put them back together if you split them like this: x^2 + x^3. They would not be "like" terms, one is squared and the other cubed. I hope this helps with your reverse engineering because I am not familiar with that method. Best Regards, Jason radyogyrl Junior Member  Posts: 6 Threads: 3 Likes Received: 0 in 0 posts Likes Given: 0 Joined: May 2007 #7 07-27-2007, 06:23 PM For this particular problem, you will square 15 which will give you 225 then mulitiply 225 by 25 which will give you 5625. I had a hard time figuring this one out when I was studying for the college math clep. I just played with it until I got the correct answer. sgloer Administrator       Posts: 975 Threads: 139 Likes Received: 137 in 48 posts Likes Given: 3 Joined: Jul 2017 #8 07-27-2007, 07:14 PM tryingtesc Wrote:You have the problem set up nicely, however, you should multiply in rows 2, 3, and 4 instead of add. You can "split" the exponent, but you then must multiply. For example: If you had x^2+3; that would equal (x^2)(x^3), the bases are the same and in multiplication you would simply add the exponents making it x^5. You could not put them back together if you split them like this: x^2 + x^3. They would not be "like" terms, one is squared and the other cubed. I hope this helps with your reverse engineering because I am not familiar with that method. Best Regards, Jason Yeah, that looks right. Thanks Jason--I couldn't remember exactly how the rule went. -- Steve Webmaster, InstantCert.com « Next Oldest | Next Newest »

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